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Transcript
Problem Set 3 Answer Key, Spring 2003 1) The following questions refer to important general features of receptor tyrosine kinase and MAPK pathways. A) Describe the structural features of a receptor tyrosine kinase. 1) extracellular ligand binding domain 2) single hydrophobic transmembrane _-helix 3) cytosolic signaling domain with a) kinase domain b) tyrosine residue(s) in the (phosphorylation lip) near the catalytic site that become phosphorylated upon ligand binding and receptor dimerization allowing full kinase function c) tyrosine residues elsewhere that are phosphorylated by the kinase domain and promote binding of other proteins B) What is the purpose of Ste5 in the yeast mating pathway? Ste5 functions as a scaffold to assemble the multiple kinase components of the pathway. This helps to ensure specificity since some of the kinases in the pathway have other targets and can mediate alternative downstream responses when not bound to Ste5. For example Ste11 is involved in both the mating and osmoregulatory pathway and binding to a scaffold ensures only the desired signal pathway is activated. C) What are some differences between the switch protein GTPases G_ and Ras? 1) Ras requires a GAP to function as an effective GTPase whereas G_ does not because it has a helical domain that serves this function. 2) Due to difference #1 G_ lasts in its GTP bound active state for a much shorter time than Ras in general though a Ras-GAP can shorten the time for Ras. 3) G_ binds directly to its receptor whereas Ras is signaled through Grb2 and Sos. 4) Ras is directly anchored to the inner membrane. 5) Ras needs a GEF 2) You are studying the response of a fibroblast cell line to treatment with fibroblast growth factor which you know binds to and acts through a receptor tyrosine kinase. The cell line grows and divides when treated with growth factor, but do not in absence of growth factor. A) You express a mutant form of the fibroblast growth factor receptor lacking its entire cytoplasmic domain to a level similar to the wild type receptor. You observe that the cells grow much more slowly than normal when treated with growth factor. Explain. The mutant receptor can bind ligand, but cannot signal. RTK’s dimerize when binding ligand and are activated by transphosphorylation. Thus complexes with one mutant and one wild type receptor will fail to signal. Only complexes of two wild type receptors will signal, and these will be present at one quarter of the normal levels so growth will be slow. B) You inject anti-Ras antibodies that prevent Ras from binding Raf into a few cells. What is the growth and division phenotype of these cells? Why? The Ras antibodies bind to Ras and prevent its function. Thus it is unable to activate the MAPK pathway, and the cells do not induce gene expression of growth and division genes. The cells will not grow even in the presence of growth factor. C) You express a constitutively active Ras that remains in its GTP bound form. You observe that the cells divide and grow even in the absence of growth factor. Explain. Ras signals to the downstream MAPK pathway whenever it is bound to GTP. Since this mutant cannot hydrolyze GTP it continually signals and the cells divide despite the lack of growth factor. D) You treat the cells with RPR-130401, a chemical that inhibits farnesyl transferase, and observe the effect on the cell growth and division. What do you see and why? Ras has a farnesylation sequence which is farnesylated by farnesyl transferase. The farnesyl group targets Ras to the membrane where it functions. Without a farnesyl group Ras will be located throughout the cytoplasm and will not be efficiently activated. Thus the cells will not grow even in the presence of growth factor. E) You combine the activated Ras mutation from part C with the RPR-130401 treatment from part D. What is the cellular growth phenotype? Despite the constitutively active Ras without the farnesylation Ras cannot efficiently signal to Raf and activate the MAPK pathway. Raf is activated primarily by being brought to the membrane so if Ras is not at the membrane this will not occur. Thus the cells will again not grow as in part D. 3) You are studying the Wg/Wnt pathway in a cell line, which responds to a Wnt peptide treatment by fifteen-fold induction of the alkaline phosphatase gene. You make several mutants of proteins in Wnt pathway components and overexpress them (20fold more than wild type) in the cell line. You observe the relative level of alkaline phosphatase mRNA before and after Wnt treatment. For the following scenarios describe the ratio of alkaline phosphatase mRNA after to before Wnt treatment. Explain. A) You overexpress a B-catenin mutant that has a serine to alanine mutation thatprevents its phosphorylation by GSK3. The mutant B-catenin cannot be inhibited by GSK3 so it will always be active and induce the expression of the alkaline phosphatase gene. Thus the ratio of alkaline phosphatase mRNA before and after Wnt treatment will be ~1. B) You overexpress a GSK3 mutant that is unable to be bound by disheveled. The mutant GSK3 cannot be inhibited by disheveled so it will always be active and phosphorylated B-catenin. Thus B-catenin will never be active and the ratio of mRNA will be ~1 since there will be little expression before or after Wnt treatment. C) You overexpress a TCF mutant lacking its domain that interacts with B-catenin. The mutant TCF cannot be bound by B-catenin so expression of alkaline phosphatase can never be turned on and the ratio of mRNA before and after treatment will be ~1. D) You overexpress a B-catenin mutant that has a serine to glutamate mutation thus mimicking its phospohorylated state. The mutant B-catenin will be unable to enter the nucleus and act to induce gene expression. However, it will not be bound by GSK3 since it is already “phosphorylated”. Thus the wild type B-catenin in the cell can still respond to Wnt treatment and the ratio of alkaline phosphatase mRNA after to before will remain at ~15. 4) Cell junctions stabilize interactions and promote local communication between adjacent cells. A) Fill in the following table for each type of cell junction: Cell junction Cell-cell or cellExtracellular ECM interaction? interaction Tight junction Cell-cell Occludin & claudin from adjacent cells Adherens junction Cell-cell E-cadherins from adjacent cells Desmosomes Cell-cell desmoglein and desmocollin from Intracellular interaction Membrane occludin – ZO-1/2/3 – actin fibers Membrane Eβcadherin – α/β catenins – actin and myosin belt Plakoglobin plaques – desmoplakin – Function Impermeable seal b/w basolateral and apical membranes Rigid cell-cell attachment Strong cellcell cadherin family) Connexin hemichannels from adjacent cells keratin filaments --- Membrane integrins – plectin – keratin filaments Membrane integrin – vinculin – actin cytoskeleton Gap junction Cell-cell Hemidesmosomes Cell-ECM Cellular integrins – ECM laminin Focal adhesion Cell-ECM Cellular integrin – ECM fibronectin attachment Movement of ions & small molecules b/w adjacent cytosols ECM attachment ECM attachment B) Specify how Ca2+ affects each of these cell junctions. Tight junctions are disrupted in the absence of Ca2+ in the medium (unknown mechanism). Adherens junctions and desmosomes are disrupted in the absence of Ca2+ in the medium since these structures contain members of the cadherin family and these are Ca2+ - dependent. The channels of gap junctions are closed in the presence of high intracellular Ca2+ concentrations. As far as we know, hemi-desmosomes and focal adhesion are not affected by Ca2+. 5) You have a mutant mouse line that cannot fight infections and die shortly thereafter. You suspect these mice have a problem in leukocyte extravasation which allows leukocytes to move from the bloodstream into the infected tissues. Name two interactions of cell adhesion molecules (CAMs) that could be defective in these mice. Indicate to what class each CAM belongs to and where each CAM is normally found. P-selectin (Selectin - endothelial cell membrane) – Sialyl Lewis-x antigen (Mucin leukocyte membrane) αLβ2 integrin (Integrin - leukocyte membrane) – ICAM-1/2 (Ig-superfamily endothelial cell membrane)